In a catalytic conversion of `N_(2)` to `NH_(3)` by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is `40 xx 10^(-3) mol L^(-1) s^(-1)`. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol `L^(-1) s^(-1)`)

Options

A. `60 xx 10^(-3)`
B. `20 xx 10^(-3)`
C. `1.200`
D. `10.3 xx 10^(-3)`

Answer

A

Explanation

`(-d(N_(2)))/(d t) = -(1)/(3)(d(H_(2)))/(d t) = (1)/(2)(d(NH_(3)))/(d t) = (3)/(2) xx 40 xx 10^(-3)`
`= 60 xx 10^(-3)`.